The given function is not a composition (other than the trivial ones). You don't use the chain rule...
–
David MitraNov 19 '11 at 17:25

1

The chain rule is for compositions of functions. If you can't write your original function as a composition, then you can't apply the chain rule. Which functions can you compose to give your original function?
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Guess who it is.Nov 19 '11 at 17:27

I do not really see any good function that I could compose the original out of. So I guess I just use no chain rule, since one has to use quotient and product rule anyway.
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Martin UedingNov 19 '11 at 22:12

Let's do it: formally substituting:
$$
f(\pi/4) = \sin\bigl(\cos(\pi/4)\bigr).
$$
To simplify this, note that you have to compute $\cos(\pi/4)$ first (you take cosine of something). $\cos(\pi/4)=\sqrt2/2$, so going back to $f$:
$$
f(\pi/4) = \sin\bigl(\cos(\pi/4)\bigr)= \sin(\sqrt2/2)\approx 0.65\,.
$$

How would you compute $g(\pi/4)$? Let's do it: formally substituting:
$$
g(\pi/4) = \cos(\pi/4) \sin(\pi/4).
$$
Here, the first thing we have to do is compute the cosine of $\pi/4$ and the sine of $\pi/4$.
We obtain
$$
g(\pi/4) = \cos(\pi/4) \sin(\pi/4) =(\sqrt2/2)\cdot(\sqrt2/2)=1/2.
$$

$f$ is a composition of functions; it's a "function of a function". That is, there is an "outer function" ($\sin$ in this case) that is evaluated at the inner function's ($\cos$ in this case) value. You use the chain rule to find the derivative of a composition:
$$
\Bigl[ f\bigr(g(x)\bigr)\Bigr]' = f'\bigr(g(x)\bigr)\cdot g'(x).
$$

$g$ is a product of two functions. For this, you of course use the product rule:
$$
\bigl[ f(x)g(x)\bigr]' =f'(x)g(x)+f(x)g'(x).
$$

When differentiating a complicated function, the first thing you should do is determine
what type it is at the "outermost level" and use the appropriate rule.

After doing some more digging, I finally came up with a way to do this with the chain rule. The point is, that my function $f$ does not solely depend on $z$ but also on $x$, yielding $f(x, z)$. If I use multivariable calculus on this one, I do get the right result.